RLC

Yes that's right Garth ,.. Thank you

Sometimes in some cases, we compare the reactance of a capacitor with resistance of a resistance like the below case

tell

When using AC-coupling between stages, the coupling capacitors can form unwanted high-pass filters, unless the reactance is negligible compared to the source resistance.

Consider the RC high-pass filter:

----C-----+--- | R |

----------+---

Transfer function = sCR / (1+sCR)

At high frequencies Vout/Vin = 1 and phase = 0; At the break-point, phase is 45 degrees and output is 3dB down; Below the break-frequency, output falls at -20dB per decade.

The break frequency is w = 1/RC

So, Vout/Vin = 1 when 1/wC

form

compared

Perhaps I should say: compared to load resistance.

Another time you might want 1/wC

yep, the resistance is the horizontal axis, the reactance is the vertical axis and the impedance is the hypotenuse.

Sounds just like the Pythagorean theorem, doesn't it....

Hi Andrew and Garth Thanks a lot for the help. Always gald to see positive guides from you all

Well my case is the Rs at source node of a nmos, and a tap capacitor is connect to the top end of Rs. So it is using 1/wc

by the way, mine is to create an oscillating circuit using nmosfet, tap capacitor, inductor as feedback. Kindly enlighthen if there is anything else to be added Thank you so much Garth and Andrew

I was trying to recall the original question.....

You had a series RCL I think that all you are seeing is that the inductive reactance and the capacitive reactance are nearly equal but opposite therefore they cancel. The circuit appears to be very much like a resistance only.

If 1/wC

Rs is the total ac resistance of cource and also resistance connecting source node and ground. I do for my assignment :)

I got my circuit oscillating but do not know how to do hand calculation to prove it .. so wish to know the reason for 1/wc

Is Rs actually Rds? (the channel resistance)

In any case, recall that you must get the proper amount of phase shift or there will be no oscillator.

Good luck!

BTW, why are you doing this?

I am sorry for my lacking in basic knowledge, why when XL and Xc cancel , we can get 180 degree phase shift?

Also the drop across the node is low meaning the current to Rs is low?

AM I right? Thank you Garth

Hi Garth

Thank you My understand is most current will be in the tank circuit no matter it is series or parallel LC tank. That means I am wrong?

Say for a nmos transistor, where its source node is connected to a tap capacitor as a feedback. Then with the tap capacitors, a inductor in series with a capacitor is connected in parallel to the earlier mentioned tap capacitors. They form a clapp LC osccilator with feeedback to the source node. For the nmos, the source node is connected to a resistor named Rs.

So this case is a parallel LC tank, is it? If yes, the condition 1/wc

Hi garth

Thank you That is really a helpful explanation I learnt about parallel LC tank only. I wonder what physics that lies behind the series LC tank to make the max current to flow to the rest of oscillator circuit without keeping most current in the LC tank(like in paralle)

By the wayI tried to email you but it failed. Is your mail box full?

Thank you so much for the help Will write again if there is any arising doubts :0 cheers

rgds and thanks jason

Do you know the memory aid for current and voltage in reactive circuits? The mnemonic is ELI the ICE man. ELI = in an L circuit the voltage is ahead of the current (in a perfect world, by 90 degrees). ICE = in a C circuit the current is ahead of the voltage ( by 90 degrees).

So there is your 180 degree shift. The definition of resonance is where Xl = Xc a condition your circuit meets.

The drop across the node is low so the current through Rs is high. Think of replacing the combined reactance with a resistor of value Z. Z would be a low value so I is high for a series LC. The story is different if LC were a parallel tank circuit. In that case the current through the tank would be low so the voltage across the tank would be high.

I hope that helps!

I think this is a point of view, whether you are looking at what happens inside the LC or what happens when you mentally replace that LC with a black box of impedance Z and consider the whole circuit.

We have been discussing what happens inside the LC and then applying that to the what happens to the remainder of the circuit. Our point of view has been shifting and this can be confusing.

In a series resonant LC, maximum current flows through LC and the rest of the oscillator circuit.

In a parallel resonant LC, maximum current is contained within the LC tank. Which implies that minimum current flows through the rest of the oscillator circuit.

You've got to mentally step back and replace the resonant LC with its resultant Z then look at the voltage across that Z

No, its just a matter of whether your looking at the whole oscillator or just the LC portion alone.

In either case, the losses are made up for by the active component or the system losses would dampen the oscillition and the circuit would ring to a stop. A mechanical equivalent would be a cymbal on a drum set, hit it once and it oscillates with a decreasing amplitude until it stops. Add an active component (a drummer that adds more energy to the system) and oscillation will continue until the input ceases, then the decay proceeds.

anything?

Thanks Garth

My doubt is " I wonder what physics that lies behind the series LC tank to make the max current to flow to the rest of oscillator circuit without keeping most current in the LC tank(like in parallel)

Jason